Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
2m(p - 6q)(p - 6q)
is eassy
5n+2n=7n If the variable is the same, n, then you can add the coefficients, 5 and 2. In General an+bn=(a+b)n FOR A CASE THAT DOES NOT WORK. 5n+2m=5n+2m You cannot combine unlike variables
n equals p over 2(like as a division problem. See, you just write n+2n=p then you do this- n+2n=p you take 2 divided by 2 and those cancel ou so if you divide by one so=ide, you have to do it to the other. so you get your answer.
N=l-m
0
Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
2m(p - 6q)(p - 6q)
is eassy
Yes.
5m+2n, n-3m, 7-3n (5m+2n)+(n-3m)+(7-3n)=2m+7
5n+2n=7n If the variable is the same, n, then you can add the coefficients, 5 and 2. In General an+bn=(a+b)n FOR A CASE THAT DOES NOT WORK. 5n+2m=5n+2m You cannot combine unlike variables
2n + 4m - 2n + m = 5m
Yes --------------------------------------------- Let n be an integer. Then 2n is an even number Let m be an integer. Then 2m is an even number and 2m + 1 is an odd number. Then: even + odd = (2n) + (2m + 1) = (2n + 2m) + 1 = 2(n + m) + 1 = 2k + 1 (where k = m + n) which is an odd number.
2n plus 4 = 2n + 4
You use the definition of even numbers which says it is a number of the form 2n for some integer n, and you take two of them, say 2n and 2m and add them and see that you have 2n+2m=2(n+m) and since n and m are integers so is n+m so lets call it p so the sum is 2p and this is even by definition of even numbers